(i) Find the value of \(k\) .

(ii) Calculate the \(y\)-coordinate of this minimum point.

Sketch the graph of \(y = f(x)\) for the domain \( - 1 \leqslant x \leqslant 3\).

(i)     \(1.25 =  - \frac{k}{{2(2)}}\)     (M1)

OR

\(f'(x) = 4x + k = 0\)     (M1)

Note: Award (M1) for setting the gradient function to zero.

\(k =  - 5\)     (A1)     (C2)

(ii)     \(2{(1.25)^2} - 5(1.25) + 4\)     (M1)

\( = 0.875\)     (A1)(ft)     (C2)

Note: Follow through from their \(k\).

[4 marks]

     (A1)(ft)(A1)(ft)     (C2)

Notes: Award (A1)(ft) for a curve with correct concavity consistent with their \(k\) passing through (0, 4).

(A1)(ft) for minimum in approximately the correct place. Follow through from their part (a).

[2 marks]

This question was not answered well at all except by the more able. Indeed, of the lower quartile of candidates, the maximum mark achieved was only 1. Of those that did make a successful attempt at the question, very few used the fact that \(1.25 = - \frac{k}{{2(2)}}\) preferring instead to differentiate and equate to zero. But such candidates were in the minority as substituting \(x = 1.25\) into the given quadratic and equating to zero produced the popular, but erroneous, answer of \(- 5.7\). Recovery was possible for the next two marks if this incorrect value had been seen to be substituted into the correct quadratic, along with \(x = 1.25\) to arrive at an answer of \(0\). This would have given (M1)(A1)(ft). However, candidates who had an answer of \(k = - 5.7\) in part (a)(i), invariably showed no working in part (ii) and consequently earned no marks here. Irrespective of incorrect working in part (a), the quadratic function clearly passes through (0, 4) and has a minimum at \(x = 1.25\). Using this information, a minority of candidates picked up at least one of the two marks in part (b).

This question was not answered well at all except by the more able. Indeed, of the lower quartile of candidates, the maximum mark achieved was only 1. Of those that did make a successful attempt at the question, very few used the fact that \(1.25 = - \frac{k}{{2(2)}}\) preferring instead to differentiate and equate to zero. But such candidates were in the minority as substituting \(x = 1.25\) into the given quadratic and equating to zero produced the popular, but erroneous, answer of \(- 5.7\). Recovery was possible for the next two marks if this incorrect value had been seen to be substituted into the correct quadratic, along with \(x = 1.25\) to arrive at an answer of \(0\). This would have given (M1)(A1)(ft). However, candidates who had an answer of \(k = - 5.7\) in part (a)(i), invariably showed no working in part (ii) and consequently earned no marks here. Irrespective of incorrect working in part (a), the quadratic function clearly passes through (0, 4) and has a minimum at \(x = 1.25\). Using this information, a minority of candidates picked up at least one of the two marks in part (b).